Question

A box contains tickets marked 1,2,3...n. A ticket is drawn at random from the box. Then the ticket is then replaced in the box and a second ticket drawn at random. Find the probabilities of the following events. a) the first ticket drawn is a number 1 and the second is the number 2. b) the numbers on the two tickets are consecutive integers, meaning the first number drawn is one less than the second number drawn. c) the second number drawn is bigger than the first number drawn.

Answer 1

The probability of specific drawing sequences from a box of numbered tickets is calculated using rules of independent events and considering the total number of possible outcomes.

Step-by-step explanation:

When analyzing the probability question involving the hypothetical drawing of a ticket from a box of numbered tickets from 1 to n, we can infer that each draw is independent due to the tickets being replaced after each draw. Therefore, the probability of each event occurring is based on the rules of independent events.

• The probability of drawing a ticket numbered '1' first and then drawing a ticket numbered '2' second is calculated by multiplying the probability of each event happening independently: P(1 then 2) = P(1) * P(2) = (1/n) * (1/n) = 1/n2.
• The probability of drawing two consecutive numbers (consecutive integers) depends on the number of possible consecutive pairs in the box, which is (n-1). Since these events are also independent, the probability will be P(consecutive pairs) = (n-1)/n2.
• To determine the probability that the second number drawn is bigger than the first number drawn, we consider all combinations of two different numbers, then subtract the number of pairs where the second number is not bigger than the first. The number of favorable outcomes is (n*(n-1))/2, and the total number of possible outcomes is n2, so the probability is P(second > first) = ((n*(n-1))/2) / n2.

Answer 2

A) 1/n²

B) (n - 1)/n²

C) Probability = (1 - (1/n))/2

Step-by-step explanation:

A) We are told that a ticket is drawn at random from the box. Thereafter, the ticket is replaced in the box and a second ticket is now drawn at random.

This means the events of the first and second tickets are independent.

Probability of first ticket = 1/n

Probability of 2nd ticket = 1/n

Probability of both together = 1/n × 1/n = 1/n²

B) The number on the 2 tickets are consecutive integers.

This means number of cases will be (n - 1)

From a above, we see that for the two tickets the denominator is n²

Thus, in this case, the probability that both that the numbers on the two tickets are consecutive integers = (n - 1)/n²

C) if we assume that the first ticket is b, then the number of correct selections for the 2nd ticket is (n - b).

Thus, number of correct selections is;

(n_Σ_k=1) (n - b) = n² - (n(n - 1)/2)

Simplifying this gives;

(2n² - n² - n)/2

>> ½(n² - n)

There are n² selections in total, thus let's factorize n² out and then the remaining part in bracket will be the probability. Thus;

>> n²(1 - (1/n))/2

Probability = (1 - (1/n))/2